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Question #41906

Difficulty: 8/10 | 🟢 Free

The Question

Let f:[246,248]→Rf: [246, 248] \rightarrow \mathbb{R} be defined by f(x)=∣x2−496x+61503∣f(x) = |x^2 - 496x + 61503|. Let SS be the set of all c∈(246,248)c \in (246, 248) such that ff is differentiable at cc and f′(c)=f(248)−f(246)2f'(c) = \frac{f(248) - f(246)}{2}. The number of elements in SS is:

Options

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    3

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Modulus Function Breakdown|Mean Value Theorem Applicability|The L'Hôpital's Rule Misuse

Why This Question Is Tricky

Common Mistakes

This question tests the strict applicability of the Mean Value Theorem (MVT) and the piecewise differentiation of modulus functions.

A common fatal error is to blindly apply MVT or differentiate the inner function without respecting the modulus, which yields c=247.5c = 247.5.

Since 247.5∈(246,248)247.5 \in (246, 248), students fall into the trap of assuming this cc is valid and answer 1.

However, MVT guarantees a solution only if the function is differentiable on the entire open interval.

Because f(x)f(x) has a 'kink' at x=247x = 247, MVT is inapplicable, and piecewise analysis reveals no valid cc exists.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterContinuity and Differentiability - Class 12
ConceptModulus Function Breakdown
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree